Question

If $\sum _{r=1}^n{\tan }^{-1}\left(\frac{x_r-x_{r-1}}{1+x_{r-1}{x}_r}\right)=$ $\sum _{r=1}^n({\tan }^{-1}{x}_r-{\tan }^{-1}{x}_{r-1})={\tan }^{-1}{x}_n-{\tan }^{-1}{x}_0$

The sum of the series ${\tan }^{-1}\frac{1}{1+1+1^2}+{\tan }^{-1}\frac{1}{1+2+2^2}+{\tan }^{-1}\left(\frac{1}{1+3+3^2}\right)+......\infty$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. None of these

Answer 1

The correct option is C $\frac{\pi }{4}$

Given

$\sum _{r=1}^n{\tan }^{-1}\left(\frac{x_r-x_{r-1}}{1+x_{r-1}{x}_r}\right)={\tan }^{-1}{x}_n-{\tan }^{-1}{x}_0$

$\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{1+r+r^2}\right)$

$=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{(r+1)-r}{1+r(1+r)}\right)$

$=\sum _{r=1}^{\infty }{\tan }^{-1}(r+1)-{\tan }^{-1}r$

$={\tan }^{-1}2-{\tan }^{-1}1+{\tan }^{-1}3-{\tan }^{-1}\mathrm{2.....}{\tan }^{-1}\infty$

$={\tan }^{-1}\infty -{\tan }^{-1}1$

$⟹\frac{\pi }{2}-\frac{\pi }{4}$

$⟹\frac{\pi }{4}$